Some Proofs on Limits and Continuity

Here are a couple of proofs that I think are nice. By nice, I mean, I like how I proved the claim because the proof is concise, clear, and rigorous.

1. Let f:R->R be a continuous function. For all r in Q, let f(r)=0.
   Claim: f(x)=0 for all x in R.
   Pf: Because f is continuous, for all c in R, for all V=N(f(c);ß), there exists U=N(c;∂) such that f(U) is a subset of V. So, V=(f(c)-ß, f(c)+ß); U=(c-∂, c+∂); and f(U) is contained in V. By the density of the rationals, there is at least one rational number in U, so f(U)=0&f(U\Q). Now, if c is rational, then V=(-ß, ß), which obviously contains 0, but must also contain f(U\Q) for all ß>0. That is, f(U\Q) is between 0 and ß, for all ß, so f(U\Q)=0. If c is irrational, then V=(f(c)-ß, f(c)+ß), which must contain 0. But if f(c)≠0, then, if ß≤f(c), 0 is not in V. So, f(c)=0. qed.
2. This one is pretty facile, but I still liked how I proved it.
   Let f:D->R, with c in D', and lim(x->c)f(x)=L>0.
   Claim: There is V=N*(c, ß): f(x)>0, for all x in R.
   Pf: Pick ß such that ß0. Then, ß>|f(x)-L| implies that f(x)-L>-ß, so f(x)>L-ß. But, by choice, L>ß>0, so L-ß>0, so f(x)>0. That is, let V be the deleted neighborhood of c, N*(c;ß), were L>ß>0. Thus, for all x in V&D, ß>|f(x)-L| implies that f(x)>L-ß<, so f(x)>0. qed.

Proof of Mathematical Principle of Induction and Other Things

Here is a proof of the Mathematical Principle of Induction.

Claim: Let P(n) be a statement for each n in N. P(n) is true for all n in N if: (a) P(1) is true, and (b) for each k in N, if P(k) is true, then P(k+1) is true.

Pf: Assume that the above conditions hold, but that there exists a number n in N such that P(n) is false. Let S be the set S={n: P(n) is false}. S is not empty by assumption, so, by the well-ordering property, there exists a least element of S, say, m. Then P(m) is false. P(1) is true by assumption, so m<1.>

This is a proof of the Archimedean Property of R.

Claim: The natural numbers N are unbounded above in the real numbers R.

Pf: By the completeness axiom, if N is bounded above, then there exists m in N such that supN=m, that is, there is a least upper bound of N. Then, m-1 is not an upper bound of N, so there exists n in N such that n>m-1. Then, n+1>m. But n+1 is in N, which contradicts m being an upper bound of N. qed.

And here's a fun little about supremums.

Claim: Given nonempty subsets A and B or R, let C denote the set C={x+y: x in A, y in B}. If A and B have supremums a and b, respectively, then supC=c=a+b.

Pf: If z is in C, then z=x+y for some x and y in A and B, respectively. Then, z=x+y≤a+b, by the definition of the supremum, so a+b is an upper bound for C. Now, choose any ∂>0. Since a is the least upper bound of A, a-∂ is not an upper bound, so there exists x in A such that a-∂

Here's one about maximums and minimums...

Claim: If S is a closed, bounded subset of R, then S has a maximum and minimum.

Pf: Let S be a closed, bounded subset of R. Then, there exists m in S such that supS=m. Because m is the least upper bound, m-∂ is in S, for all ∂>0, but m+∂ is in R\S; that is, for all ∂>0, N(m,∂)&S and N(m,∂)&R\S are nonempty, so m is a boundary point of S. S is closed, so m is in S. Similarly for a minimum. qed.

And, finally, one about convergent sequences. Enjoy.

Claim: Every convergent sequence is bounded.

Pf: Let the sequence (Sn) converge to s. and let (Sn) be unbounded. Then, letting ∂=1, we get N in R such that n>N and 1>|sn-s|. Then, the triangle inequality implies that |sn|<|s|+1. Letting M=max{|s1|,|s2|,...,|sN|, |s|+1}, then |sn|<=M for all n in N, so (Sn) is bounded.

Ok, that's all for now.